Question

For a weak acid with a $pK_a$ value of 6.0, the ratio of acid to salt at pH 5.0 will be:

• A. 1.0
• B. 10.0
• C. 6.0
• D. 100.0

Hint:

Rule of Thumb:
If $pH = pK_a$, ratio is 1:1.
If $pH$ is 1 unit below $pK_a$, Acid:Salt is 10:1.
If $pH$ is 2 units below $pK_a$, Acid:Salt is 100:1.
Always check if the question asks for Salt:Acid or Acid:Salt!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks to calculate the quantitative ratio of the undissociated acid form to the salt (conjugate base) form for a specific weak acid given the $pH$ and $pK_a$.
This is a standard calculation involving buffer systems and acid-base equilibrium.
Key Formula or Approach:
The Henderson-Hasselbalch equation is used to relate $pH$, $pK_a$, and the concentrations of the salt and acid:
\[ pH = pK_a + \log_{10} \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \]

Step 2: Detailed Explanation:

• Substituting the Values: Given $pH = 5.0$ and $pK_a = 6.0$.
\[ 5.0 = 6.0 + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \]

• Isolating the Logarithm: Subtract 6.0 from both sides:
\[ 5.0 - 6.0 = \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \]
\[ -1.0 = \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \]

• Removing the Logarithm: Take the antilog (base 10) of both sides:
\[ 10^{-1.0} = \frac{[\text{Salt}]}{[\text{Acid}]} \]
\[ 0.1 = \frac{[\text{Salt}]}{[\text{Acid}]} \]

• Finding the Inverse Ratio: The question specifically asks for the ratio of Acid to Salt:
\[ \frac{[\text{Acid}]}{[\text{Salt}]} = \frac{1}{0.1} = 10.0 \]

• Conceptual check: Since the $pH$ (5.0) is lower than the $pK_a$ (6.0), the solution is more acidic than the point where acid and salt are equal. Therefore, the acid form must predominate over the salt form. A ratio of 10.0 for Acid:Salt makes physical sense.

Step 3: Final Answer:
The ratio of acid to salt at $pH$ 5.0 for a weak acid with $pK_a$ 6.0 is 10.0.